Problem: $g(x)=x^2+3x-4$ 1) What are the zeros of the function? Write the smaller $x$ first, and the larger $x$ second. $\text{smaller }x=$
Solution: To find the zeros of the function, we need to solve the equation $g(x)=0$. We can do that by factoring $g(x)$. $\begin{aligned} x^2+3x-4&=0 \\\\ (x-1)(x+4)&=0 \\\\ x-1=0&\text{ or }x+4=0 \\\\ x={1}&\text{ or }x={-4} \end{aligned}$ There are many ways to find the vertex. We will do it by using the fact that the $x$ -coordinate of the vertex is exactly between the two zeros. $\begin{aligned} \text{vertex's }x\text{-coordinate}&=\dfrac{({1})+({-4})}{2} \\\\ &={-\dfrac32} \end{aligned}$ Now we can find the vertex's $y$ -coordinate by evaluating $g\left({-\dfrac32}\right)$ : $\begin{aligned} g\left({-\dfrac32}\right)&=\left({-\dfrac32}\right)^2+3\left({-\dfrac32}\right)-4 \\\\ &=\dfrac94-\dfrac92-4 \\\\ &=-\dfrac{25}{4} \end{aligned}$ In conclusion, $\begin{aligned} \text{smaller }x&=-4 \\\\ \text{larger }x&=1 \end{aligned}$ The vertex of the parabola is at $\left(-\dfrac32,-\dfrac{25}{4}\right)$